Type-curve matching

Quick refresher

In logarithms, the change from real quantities to dimensionless ones is a simple shift. A record that obeys the model is a copy of one of the type curves, shifted parallel to the axes.
The shift along the pressure axis (the pressure match) gives the flow capacity $kh$ . The shift along the time axis (the time match) gives the wellbore storage coefficient $C$ . The porosity, the compressibility of the reservoir and the wellbore radius are not in the shifts.
The parameter of the curve, $C_De^{2S}$ , is the dimensionless wellbore storage coefficient taken at the effective wellbore radius $r_we^{-S}$ (the radius of an ideal well without skin that would behave like the real well). The skin is half the natural logarithm of the ratio of the parameter to $C_D$ . So an error of a factor of 2 in the parameter changes the skin by only 0.35.
We start the match with the unit slope, which links the two shifts. Then we move the record along it until the transition section and the late points lie on one of the curves.
The match is reliable only if the record has reached radial flow. If the record ended in the transition section, neighbouring curves describe it almost equally well, and the permeability and the skin are uncertain. $C$ is still well known.
We check the result independently: $C$ by the unit slope, and $kh$ and the skin by a semilog analysis of the points that the match assigned to radial flow.

Key relationships

\lg p_D = \lg\Delta p + \lg\dfrac{2\pi kh}{qB\mu},\qquad \lg\dfrac{t_D}{C_D} = \lg t + \lg\dfrac{2\pi kh}{\mu C}

shifts between the plot of the record and the type-curve plot (SI)

kh = 1.866\,qB\mu\left(\dfrac{p_D}{\Delta p}\right)_M

pressure match: m³/d, mPa·s, MPa, mD·m (in oilfield units the constant is 141.2)

C = 0.02232\,\dfrac{kh}{\mu}\left(\dfrac{\Delta t}{t_D/C_D}\right)_M

time match: m³/MPa, mD·m, mPa·s, h (oilfield: 0.000295)

C_D = \dfrac{C}{2\pi\phi c_t h r_w^2},\qquad S = \tfrac12\ln\dfrac{\left(C_De^{2S}\right)_M}{C_D}

the skin from the parameter of the curve; the index $M$ means the values at the match point

Common pitfalls

Choosing the curve by the early points and not checking the late ones. In the transition section, neighbouring curves are almost the same.
Turning the plot when we lay it over the curves, or drawing it with a different length of the logarithmic cycle.
Setting the moment of start-up or shut-in inexactly. The early points then shift in time, and the unit slope is distorted.
Matching a record to curves for a constant $C$ when the wellbore storage coefficient or the skin changes during the record.
Taking a good fit as proof of the model. Other reservoir models give similar records too.

After this lesson you can

explain why the record $\Delta p(t)$ on a log–log plot repeats the shape of the dimensionless curve of the model, and what the shifts along the axes contain
describe the family of type curves for a well with wellbore storage and skin in the coordinates $p_D$ against $t_D/C_D$ , and the meaning of the parameter $C_De^{2S}$
do a match, and from the match point calculate the flow capacity, the permeability, the wellbore storage coefficient $C$ , $C_D$ and the skin in metric and oilfield units
check the result with the unit-slope section (the early straight line of slope 1) and the semilog straight line, and tell which points belong to radial flow
name the causes of an ambiguous match and the errors it leads to: a short record, a time count that starts at the wrong moment, a changing wellbore storage coefficient, an unsuitable model

The shape of the curve and the shifts along the axes

Dimensionless variables (dimensionless form) make the solution common to all reservoirs. We measure the pressure in “units” of $qB\mu/(2\pi kh)$ and the time in units tied to the reservoir and the wellbore. For a well with wellbore storage, a convenient dimensionless time is $t_D/C_D$ (Wellbore storage). If we write the change to dimensionless quantities in logarithms, multiplying by a constant factor turns into adding a constant number:

The link between the plot of the record and the type-curve plot, in logarithms

\lg p_D = \lg\Delta p + \lg\frac{2\pi kh}{qB\mu},\qquad \lg\frac{t_D}{C_D} = \lg t + \lg\frac{2\pi kh}{\mu C}

where:

$p_D$ dimensionless change of pressure, $2\pi kh\,\Delta p/(qB\mu)$ , —
$\Delta p$ change of the bottomhole pressure since start-up ( $p_i - p_{wf}$ ) or shut-in ( $p_{ws} - p_{wf}(\Delta t = 0)$ ), Pa
$t_D/C_D$ dimensionless time divided by the dimensionless wellbore storage coefficient: $2\pi kh\,t/(\mu C)$ , —
$t$ time since start-up or shut-in (for a build-up, $\Delta t$ ), s
$q,\ B,\ \mu$ change of the rate at the wellhead at standard conditions, formation volume factor (reservoir volume per unit of surface volume), viscosity, m³/s, m³/m³, Pa·s
$k,\ h$ permeability and thickness of the reservoir, m², m
$C$ wellbore storage coefficient, m³/Pa

Units: SI; the forms below are in oilfield units

Holds when

A start-up from a state in which the pressure in the reservoir has equalised, or a shut-in after production at a constant rate. For a pressure build-up curve, the shut-in time is much shorter than the producing time of the well.
The record obeys the model of the type curves: a homogeneous infinite reservoir, a constant $C$ and a constant skin.

The second term in each formula is a constant number. So on a plot with two logarithmic scales the record $\Delta p(t)$ is the same curve as $p_D(t_D/C_D)$ , only shifted: vertically by $\lg[2\pi kh/(qB\mu)]$ , and horizontally by $\lg[2\pi kh/(\mu C)]$ . The shape of the curve depends only on the model and its parameter. The shifts contain $kh$ and $C$ . The porosity, the compressibility of the reservoir and the wellbore radius cancelled out of $t_D/C_D$ .

In other unit systems (2)

Metric oilfield units

\frac{p_D}{\Delta p} = \frac{kh}{1.866\,qB\mu},\qquad \frac{t_D/C_D}{t} = 0.02232\,\frac{kh}{\mu C}

$q$ in m³/d at standard conditions, $\mu$ in mPa·s, $k$ in mD, $h$ in m, $\Delta p$ in MPa, $t$ in h, $C$ in m³/MPa.

$1.866 = 1/(2\pi\cdot0.0853)$ , where $0.0853$ is the constant of Darcy's law in these units (Skin factor); $0.02232 = 2\pi\cdot3.553\cdot10^{-3}$ , as in the lesson Wellbore storage.

Oilfield units

\frac{p_D}{\Delta p} = \frac{kh}{141.2\,qB\mu},\qquad \frac{t_D/C_D}{t} = 0.000295\,\frac{kh}{\mu C}

$q$ in STB/d (stock-tank barrels of oil per day), $\mu$ in cP, $k$ in mD, $h$ in ft, $\Delta p$ in psi, $t$ in h, $C$ in bbl/psi.

$141.2 = 1/(2\pi\cdot1.127\cdot10^{-3})$ ; $0.000295 = 2\pi\cdot2.637\cdot10^{-4}/5.615$ (5.615 is the number of cubic feet in a barrel).

The family of curves for a well with wellbore storage and skin

The model is a homogeneous infinite reservoir and a well with a constant wellbore storage coefficient and a constant skin. Its dimensionless solution depends on two numbers, $C_D$ and $S$ . If we plot $p_D$ against $t_D$ , each pair needs its own curve, and it is awkward to fit two parameters at once. Two things help: another dimensionless time, $t_D/C_D$ , and a single parameter, $C_De^{2S}$ .

Why one parameter is enough. A well with a skin $S$ gives almost the same bottomhole pressure as a well without skin that has the effective radius $r_we^{-S}$ (Skin factor). The dimensionless wellbore storage coefficient taken at this radius is $C_De^{2S}$ : the square of the radius in the denominator of $C_D$ gets smaller by a factor of $e^{2S}$ . In the variables of this “effective” well, the problem has one parameter, and it is the number written on the curves. The curves of this lesson were calculated in this way.

The parameter of the type curves and the two asymptotes of the family

C_De^{2S} = \frac{C}{2\pi\phi c_t h\left(r_we^{-S}\right)^2};\qquad p_D = \frac{t_D}{C_D}\ \text{(wellbore)},\qquad p_D = \frac12\left[\ln\frac{t_D}{C_D} + 0.809 + \ln\left(C_De^{2S}\right)\right]\ \text{(reservoir)}

where:

$C_De^{2S}$ parameter of the type curve, —
$C$ wellbore storage coefficient, m³/MPa
$\phi,\ c_t,\ h$ porosity, total compressibility, thickness of the reservoir, —, 1/MPa, m
$r_w,\ S$ wellbore radius and skin factor; $r_we^{-S}$ is the effective radius, m, —
$p_D,\ t_D/C_D$ dimensionless change of pressure and dimensionless time, —

Units: dimensionless ( $C$ and $c_t$ are given per the same unit of pressure); natural logarithm

Holds when

A homogeneous infinite reservoir; constant $C$ and $S$ .
The first asymptote is the section of pure wellbore storage. The second is radial flow, which starts after the transition section (how long wellbore storage lasts).
Replacing the skin with an effective radius is exact for radial flow and almost exact in the transition section, if $C_D$ is not too small (advanced section).

All the curves start on one unit-slope line: while only the wellbore works, the reservoir does not matter. Then each curve leaves it and goes to its own logarithmic asymptote (a line that the curve approaches more and more closely). The larger the parameter, the longer the curve follows the unit slope, the sharper it then turns, and the higher its asymptote lies. The second asymptote is the same pressure function $\tfrac12(\ln t_D + 0.809) + S$ (the pressure function), written with $t_D/C_D$ .

CalculatedCalculated family of type curves for a well with wellbore storage and skin in an infinite reservoir. Both scales are logarithmic. From bottom to top, the curves have the parameter

C_De^{2S}

from

10^0

10^{10}

in steps of two orders of magnitude (each step is a factor of 100). The dashed straight line is the unit slope

p_D = t_D/C_D

, common to all the curves.

The parameter

C_De^{2S}

for different

C_D

and skin

$C_D$: 10
$S = -2$: $0.18$
$S = 0$: $10$
$S = 3$: $4034$
$S = 8$: $8.9\cdot10^{7}$

$C_D$: 100
$S = -2$: $1.8$
$S = 0$: $100$
$S = 3$: $4.0\cdot10^{4}$
$S = 8$: $8.9\cdot10^{8}$

$C_D$: 1000
$S = -2$: $18$
$S = 0$: $1000$
$S = 3$: $4.0\cdot10^{5}$
$S = 8$: $8.9\cdot10^{9}$

$C_D$	$S = -2$	$S = 0$	$S = 3$	$S = 8$
10	$0.18$	$10$	$4034$	$8.9\cdot10^{7}$
100	$1.8$	$100$	$4.0\cdot10^{4}$	$8.9\cdot10^{8}$
1000	$18$	$1000$	$4.0\cdot10^{5}$	$8.9\cdot10^{9}$

A $C_D$ of about 10 belongs to shut-in at the bottom of the well, about 100 to a wellbore filled with liquid, and about 1000 to free gas in the wellbore (Wellbore storage). The parameter grows as $e^{2S}$ : each unit of skin multiplies it by 7.4. So the position of the curve in the family is set above all by the skin.

What changes if…

…the skin is larger by one: the parameter is 7.4 times larger, and the radial asymptote is higher by one unit of $p_D$ .
… $C$ is twice as large: everything connected with the wellbore comes twice as late in the record. The parameter is twice as large, and at the same $t_D/C_D$ the asymptote is higher by $\tfrac12\ln 2 = 0.35$ . But the same moment of the record corresponds to a twice smaller $t_D/C_D$ . So the radial-flow pressure does not depend on $C$ .
… $kh$ is twice as large: the record shifts along both the pressure axis and the time axis, and the curve stays the same. The permeability is not in the parameter.
…we know the porosity, the compressibility or the wellbore radius only roughly: the shifts do not change. What changes is the calculated $C_D$ , and with it the skin.

How we match a record to the curves

Matching means finding two shifts and one parameter. By hand, people did it in the steps given below. Programs do the same by fitting (section “Matching today”). The main condition is that the type-curve plot and the plot of the record are drawn with the same length of the logarithmic cycle, and that the axes stay parallel when we lay one over the other.

Order of matching

1
Draw the plot of the record
Plot $\Delta p$ against $\Delta t$ on log–log axes, with the same cycle length as the type curves. For a build-up, $\Delta p$ is the pressure rise since shut-in.
2
Lay it over without turning
Place the plot of the record over the family and move it only up and down and right and left.
3
Match the unit slope
Put the early points on the unit-slope line. This links the two shifts: from here on we move the plot only along this line.
4
Choose the curve
Move along the unit slope and find the curve on which the transition section and the late points lie. If needed, interpolate between neighbouring curves.
5
Read the match point
Choose a convenient point and write down its coordinates in both systems: $p_D$ and $t_D/C_D$ on the type-curve plot, $\Delta p$ and $\Delta t$ on the plot of the record. Write down the parameter of the curve.
6
Calculate the parameters
Find $kh$ , $C$ , $C_D$ and the skin by the formulas below.
7
Check
Compare $C$ with the unit-slope section and with the design of the well. Find $kh$ and the skin by a semilog analysis of the points that the match assigned to radial flow. Make sure that the late points are not distorted by boundaries.

Pressure match: the flow capacity

kh = \frac{qB\mu}{2\pi}\left(\frac{p_D}{\Delta p}\right)_M

where:

$\left(p_D/\Delta p\right)_M$ pressure match: the dimensionless pressure of the match point divided by the change of pressure that coincided with it, 1/Pa
$kh$ flow capacity, m³
$q,\ B,\ \mu$ rate at standard conditions (after start-up or before shut-in), formation volume factor, viscosity, m³/s, m³/m³, Pa·s

Units: SI

Holds when

The record is matched to the curve as a whole, including the radial-flow section.
The rate is constant. For a build-up, the shut-in time is small compared with the producing time.

The higher the points of the record lie relative to the curve (a larger $\Delta p$ at the same $p_D$ ), the lower the flow capacity. In meaning this is the same as the slope of the semilog straight line: in the radial section $p_D$ grows by 1.151 per decimal cycle.

In other unit systems (2)

Metric oilfield units

kh = 1.866\,qB\mu\left(\frac{p_D}{\Delta p}\right)_M

$q$ in m³/d, $\mu$ in mPa·s, $\Delta p$ in MPa; $kh$ in mD·m.

$1.866 = 1/(2\pi\cdot0.0853)$ : the same constant as in the formula for the skin pressure drop.

Oilfield units

kh = 141.2\,qB\mu\left(\frac{p_D}{\Delta p}\right)_M

$q$ in STB/d, $\mu$ in cP, $\Delta p$ in psi; $kh$ in mD·ft.

$141.2 = 1/(2\pi\cdot1.127\cdot10^{-3})$ .

Time match: the wellbore storage coefficient

C = \frac{2\pi kh}{\mu}\left(\frac{\Delta t}{t_D/C_D}\right)_M

where:

$\left(\Delta t/(t_D/C_D)\right)_M$ time match: the time of the record that coincided with the dimensionless time of the match point, divided by this dimensionless time, s
$C$ wellbore storage coefficient, m³/Pa
$kh,\ \mu$ flow capacity (from the pressure match) and viscosity, m³, Pa·s

Units: SI

Holds when

$kh$ was found from the pressure match.
Time is counted exactly from the moment of start-up or shut-in.

For this form of the curves, the time match gives not the permeability but the wellbore storage coefficient. The same $C$ must come from the unit-slope section, and we can compare it with the design of the well and its contents.

In other unit systems (2)

Metric oilfield units

C = 0.02232\,\frac{kh}{\mu}\left(\frac{\Delta t}{t_D/C_D}\right)_M

$kh$ in mD·m, $\mu$ in mPa·s, $\Delta t$ in h; $C$ in m³/MPa.

$0.02232 = 2\pi\cdot3.553\cdot10^{-3}$ , where $3.553\cdot10^{-3}$ converts time in hours into dimensionless time.

Oilfield units

C = 0.000295\,\frac{kh}{\mu}\left(\frac{\Delta t}{t_D/C_D}\right)_M

$kh$ in mD·ft, $\mu$ in cP, $\Delta t$ in h; $C$ in bbl/psi.

$0.000295 = 2\pi\cdot2.637\cdot10^{-4}/5.615$ .

Dimensionless wellbore storage coefficient and skin from the parameter of the curve

C_D = \frac{C}{2\pi\phi c_t h r_w^2},\qquad S = \frac12\ln\frac{\left(C_De^{2S}\right)_M}{C_D}

where:

$\left(C_De^{2S}\right)_M$ parameter of the curve on which the record lies, —
$C_D$ dimensionless wellbore storage coefficient, —
$C$ wellbore storage coefficient from the time match, m³/MPa
$\phi,\ c_t,\ h,\ r_w$ porosity, total compressibility, thickness of the reservoir, wellbore radius, —, 1/MPa, m, m
$S$ skin factor, —

Units: metric oilfield units or SI: $C$ and $c_t$ are given per the same unit of pressure; natural logarithm

Holds when

$\phi$ , $c_t$ , $h$ and $r_w$ are known independently: $C_D$ depends on them, and so does the skin.
The skin that we find refers to the wellbore radius that we put into $C_D$ .

We read the parameter of the curve on a logarithmic scale, and the skin depends on its logarithm. An error of a factor of 2 in the parameter changes the skin by $\tfrac12\ln 2 = 0.35$ , and an error of a factor of 10 changes it by 1.15.

In other unit systems (1)

Oilfield units

C_D = \frac{0.894\,C}{\phi c_t h r_w^2}

$C$ in bbl/psi, $c_t$ in 1/psi, $h$ and $r_w$ in ft.

$0.894 = 5.615/(2\pi)$ : the numerator is in barrels and the denominator is in cubic feet (as in the lesson Wellbore storage).

Example: matching a pressure build-up record

An oil well produced for about four months (3000 h) at 30 m³/d. Then we shut it in at the wellhead and recorded the bottomhole pressure for four days. There is free gas in the wellbore, so wellbore storage lasts for hours. Below are an assumed record and its match with the type curves.

Rise of the bottomhole pressure after shut-in (assumed data)

$\Delta t$: 10 s
$\Delta p$ , MPa: 0.052
$\Delta p/\Delta t$ , MPa/h: 18.7

$\Delta t$: 20 s
$\Delta p$ , MPa: 0.103
$\Delta p/\Delta t$ , MPa/h: 18.5

$\Delta t$: 40 s
$\Delta p$ , MPa: 0.203
$\Delta p/\Delta t$ , MPa/h: 18.3

$\Delta t$: 1.5 min
$\Delta p$ , MPa: 0.442
$\Delta p/\Delta t$ , MPa/h: 17.7

$\Delta t$: 3 min
$\Delta p$ , MPa: 0.838
$\Delta p/\Delta t$ , MPa/h: 16.8

$\Delta t$: 6 min
$\Delta p$ , MPa: 1.521
$\Delta p/\Delta t$ , MPa/h: 15.2

$\Delta t$: 12 min
$\Delta p$ , MPa: 2.553
$\Delta p/\Delta t$ , MPa/h: 12.8

$\Delta t$: 30 min
$\Delta p$ , MPa: 4.165
$\Delta p/\Delta t$ , MPa/h: 8.33

$\Delta t$: 1 h
$\Delta p$ , MPa: 5.035
$\Delta p/\Delta t$ , MPa/h: 5.04

$\Delta t$: 2 h
$\Delta p$ , MPa: 5.470
$\Delta p/\Delta t$ , MPa/h: 2.73

$\Delta t$: 4 h
$\Delta p$ , MPa: 5.738
$\Delta p/\Delta t$ , MPa/h: 1.43

$\Delta t$: 8 h
$\Delta p$ , MPa: 5.965
$\Delta p/\Delta t$ , MPa/h: 0.75

$\Delta t$: 16 h
$\Delta p$ , MPa: 6.178
$\Delta p/\Delta t$ , MPa/h: 0.39

$\Delta t$: 24 h
$\Delta p$ , MPa: 6.300
$\Delta p/\Delta t$ , MPa/h: 0.26

$\Delta t$: 48 h
$\Delta p$ , MPa: 6.503
$\Delta p/\Delta t$ , MPa/h: 0.14

$\Delta t$: 72 h
$\Delta p$ , MPa: 6.620
$\Delta p/\Delta t$ , MPa/h: 0.09

$\Delta t$: 96 h
$\Delta p$ , MPa: 6.703
$\Delta p/\Delta t$ , MPa/h: 0.07

$\Delta t$	$\Delta p$ , MPa	$\Delta p/\Delta t$ , MPa/h
10 s	0.052	18.7
20 s	0.103	18.5
40 s	0.203	18.3
1.5 min	0.442	17.7
3 min	0.838	16.8
6 min	1.521	15.2
12 min	2.553	12.8
30 min	4.165	8.33
1 h	5.035	5.04
2 h	5.470	2.73
4 h	5.738	1.43
8 h	5.965	0.75
16 h	6.178	0.39
24 h	6.300	0.26
48 h	6.503	0.14
72 h	6.620	0.09
96 h	6.703	0.07

$\Delta p = p_{ws} - p_{wf}(\Delta t = 0)$ . The data were calculated from a model of a homogeneous infinite reservoir with wellbore storage and skin, and rounded to 0.001 MPa. The parameters of the model are given at the end of the example. The almost constant $\Delta p/\Delta t$ in the first tens of seconds is a sign of the unit-slope section.

CalculatedTop: the record from the table on log–log axes. Bottom: the same record shifted onto the family of type curves (grey: the parameter from

10^0

10^{10}

in steps of two orders of magnitude; orange: the fitted curve,

C_De^{2S} = 1.4\cdot10^6

). The length of the logarithmic cycle is the same on both plots, so the points keep their positions relative to each other. The match point is marked:

p_D = 1

t_D/C_D = 100

at the bottom, and the

\Delta t

and

\Delta p

that coincided with it at the top.

Worked examplePermeability, wellbore storage coefficient and skin from the match

Given (assumed data)

Well and reservoir: $q$ = 30 m³/d; $B$ = 1.2; $\mu$ = 1.3 mPa·s; $h$ = 10 m; $\phi$ = 0.17; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m
Producing time before shut-in: 3000 h, much longer than the record
Result of the match: the record lay on the curve $C_De^{2S} = 1.4\cdot10^{6}$ ; the point $p_D = 1$ , $t_D/C_D = 100$ coincided with $\Delta p$ = 0.583 MPa and $\Delta t$ = 3.10 h

Solution

Pressure match: $qB\mu = 30\cdot1.2\cdot1.3 = 46.8$ ; $kh = 1.866\cdot46.8\cdot(1/0.583) = 149.8$ mD·m; $k = 149.8/10 = 14.98$ mD.
Time match: $C = 0.02232\cdot\dfrac{149.8}{1.3}\cdot\dfrac{3.10}{100} = 0.0797$ m³/MPa.
The same in oilfield units (1 m³ = 6.2898 bbl, 1 MPa = 145.04 psi, 1 m = 3.2808 ft): $q$ = 188.7 STB/d, $\Delta p_M$ = 84.56 psi; $kh = 141.2\cdot188.7\cdot1.2\cdot1.3/84.56 = 492$ mD·ft, that is, 149.8 mD·m; $C = 0.000295\cdot(492/1.3)\cdot(3.10/100) = 0.00346$ bbl/psi, that is, 0.0797 m³/MPa.
Dimensionless coefficient: $2\pi\phi c_thr_w^2 = 2\pi\cdot0.17\cdot1.5\cdot10^{-3}\cdot10\cdot0.1^2 = 1.602\cdot10^{-4}$ m³/MPa; $C_D = 0.0797/(1.602\cdot10^{-4}) = 498$ .
Skin: $S = \tfrac12\ln\dfrac{1.4\cdot10^{6}}{498} = \tfrac12\ln 2811 = 3.97$ .
Check by the unit-slope section: at 10 s, $\Delta p$ = 0.052 MPa, and $C = \dfrac{qB\,\Delta t}{24\,\Delta p} = \dfrac{36\cdot(10/3600)}{24\cdot0.052} = 0.0801$ m³/MPa. This is practically the same as from the match.
Where radial flow starts: after the match, the points from 4 h on lie on the radial asymptote of the fitted curve with a deviation of less than 1%. We take them for the semilog check, in the next example.

Answer:

kh \approx 149.8

mD·m,

k \approx 14.98

mD;

C \approx 0.0797

m³/MPa (

C_D \approx 498

); the skin is about 3.97.

What it means. We calculated the record from a model with $k$ = 15 mD, $S$ = 4 and $C$ = 0.08 m³/MPa. The match returned them with errors of 0.1% in permeability, 0.3% in $C$ and 0.03 in skin. The errors come from the rounding of the measurements, and from the fact that the record is a build-up and not a drawdown: by the end of the record, the pressure rise is 0.14% smaller than on the start-up curve. The wellbore storage coefficient is several times larger than a wellbore filled with oil alone would give (in the lesson Wellbore storage it is 0.013 m³/MPa for 8 m³ of wellbore). The reason is free gas in the wellbore. The data in this example are assumed for illustration.

Worked exampleCheck with the Horner line

Given (assumed data)

Points of radial flow: by the match, from 4 h on; we take two points far apart: 24 h (6.300 MPa) and 96 h (6.703 MPa)
Producing time; $qB\mu$: 3000 h; 46.8
Reservoir: $h$ = 10 m; $\phi$ = 0.17; $\mu$ = 1.3 mPa·s; $c_t = 1.5\cdot10^{-3}$ MPa⁻¹; $r_w$ = 0.1 m

Solution

Horner ratios: $(3000 + 24)/24 = 126$ and $(3000 + 96)/96 = 32.25$ ; the difference of their base-10 logarithms is 0.59184.
Slope: $m = (6.703 - 6.300)/0.59184 = 0.6809$ MPa per cycle; $kh = 2.149\cdot46.8/0.6809 = 147.7$ mD·m; $k = 14.77$ mD.
Pressure rise on the line 1 h after shut-in: $\Delta p_{1\text{h}} = 6.300 - 0.6809\cdot(\lg 3001 - \lg 126) = 5.362$ MPa.
Skin: $\lg\dfrac{k}{\phi\mu c_tr_w^2} = 6.649$ ; $S = 1.151\,(5.362/0.6809 - 6.649 + 2.098) = 3.83$ .

Answer: By the Horner line,

k \approx 14.77

mD and

S \approx 3.83

. The match gave 14.98 mD and 3.97.

What it means. Two independent methods agree. This is the main argument for the interpretation. The small difference is natural. By 24 h the pressure already lies on the radial asymptote to within a per cent, but the slope of the curve is still a little larger than in radial flow. So the line through 24 and 96 h is 1.6% steeper than the model, and the permeability from it is lower. The match uses the whole record, and here it came closer to the parameters of the model. The skin formula is from the lesson Pressure build-up analysis. The correction for the finite producing time, $\lg[(t_p + 1)/t_p]$ , is negligible at 3000 h. The data in this example are assumed for illustration.

What can go wrong

Matching is a fit, by eye or by a program, and it has weak points. The main one is ambiguity. If the record has not reached radial flow, different pairs of “permeability and skin” describe it almost equally well. To see this, we cut the record of the example at 1 h and match it to three curves.

CalculatedThe record of the example, cut at 1 h (the points), matched to three curves:

C_De^{2S} = 10^5

(turquoise), the best one for the first hour (dark) and

10^7

(orange). Up to 1 h all three pass through the points to within a fraction of a per cent, and later they diverge. The grey dashed line is the continuation of the record, which the interpreter did not yet have. The time scale is logarithmic and the pressure scale is ordinary.

Worked exampleA short record: three almost equal answers

Given (assumed data)

Record: the same measurements, but only up to 1 h: 9 points
Curves: $C_De^{2S} = 10^5$ ; the best one for these points; $C_De^{2S} = 10^7$

Solution

The curve $10^5$ : the match gives $k$ = 12.5 mD, $C$ = 0.0795 m³/MPa, $S$ = 2.7; the root-mean-square deviation of the points from the curve (a kind of average gap) is 0.24%.
The best curve, $C_De^{2S} = 4.9\cdot10^{5}$ : $k$ = 14.0 mD, $C$ = 0.0797 m³/MPa, $S$ = 3.5; deviation 0.13%.
The curve $10^7$ : $k$ = 16.8 mD, $C$ = 0.0800 m³/MPa, $S$ = 5.0; deviation 0.31%.
For comparison, the whole record up to 96 h: on the curves $10^5$ and $10^7$ the deviation is 1.14% and 0.66%, against 0.13% for the best one.

Answer: From the first hour, the permeability is known only within 12.5–16.8 mD, and the skin from 2.7 to 5.0. The wellbore storage coefficient is known well: 0.0795–0.0800 m³/MPa.

What it means. All three matches differ from the points by less than half a per cent. On a log–log plot such differences cannot be seen, and the record of a real well usually differs from the ideal model of the wellbore by more. Even the best curve for the first hour did not coincide with the model from which the points were calculated: the rounding of the measurements decided its choice. The early points reliably give only $C$ : on the unit-slope section the reservoir is “not visible”. The permeability and the skin compensate each other: a more permeable reservoir with a larger skin gives almost the same transition. Only radial flow lets us tell them apart. In the full record, the wrong curves deviate from the points 5 to 9 times more. The data in this example are assumed for illustration.

From this comes a rule for planning: after wellbore storage ends, we must record at least one logarithmic cycle of radial flow (how long wellbore storage lasts). If that is impossible, we give the result of the match as a range, not as one number, and we check it against independent data: the permeability measured on core samples, and the neighbouring wells. Matching the pressure and its logarithmic derivative (its slope on a semilog plot) at the same time greatly reduces the ambiguity. On the derivative, the curves of neighbouring parameters differ much more clearly (Diagnosis with the pressure derivative).

Signs of a poor match

What we see: The early points lie on a line with a slope below or above 1
Possible cause: an inexact moment of start-up or shut-in; a changing $C$
What to do: refine the moment from the record of the gauge (Wellbore storage); see Nonideal wellbore storage, well clean-up and changing skin

What we see: The early points move from curve to curve
Possible cause: the wellbore storage coefficient or the skin changes, for example during well clean-up
What to do: do not match to a single curve; Nonideal wellbore storage, well clean-up and changing skin

What we see: The late points of a build-up go above the curve
Possible cause: a sealing boundary: a fault
What to do: A sealing fault near the well: the method of images

What we see: The late points of a build-up flatten below the curve
Possible cause: a closed area, or a short producing time before shut-in
What to do: Build-up in a bounded reservoir and average reservoir pressure, Effect of producing time on build-up analysis

What we see: The curve does not describe the transition section, though the start and the end fit well
Possible cause: another model: a hydraulic fracture, dual porosity
What to do: try other models; Diagnosis with the pressure derivative

What we see	Possible cause	What to do
The early points lie on a line with a slope below or above 1	an inexact moment of start-up or shut-in; a changing $C$	refine the moment from the record of the gauge (Wellbore storage); see Nonideal wellbore storage, well clean-up and changing skin
The early points move from curve to curve	the wellbore storage coefficient or the skin changes, for example during well clean-up	do not match to a single curve; Nonideal wellbore storage, well clean-up and changing skin
The late points of a build-up go above the curve	a sealing boundary: a fault	A sealing fault near the well: the method of images
The late points of a build-up flatten below the curve	a closed area, or a short producing time before shut-in	Build-up in a bounded reservoir and average reservoir pressure, Effect of producing time on build-up analysis
The curve does not describe the transition section, though the start and the end fit well	another model: a hydraulic fracture, dual porosity	try other models; Diagnosis with the pressure derivative

On a log–log plot each next cycle of time takes as much room as the previous one, though it lasts ten times longer, and late deviations from the curve look small. It is easier to look for the signs of boundaries on a semilog plot and on a plot with ordinary axes.

Matching today

While families of curves were printed on paper, people put the record on transparent tracing paper with the same logarithmic grid, and slid it by hand over the type-curve chart. Today programs do the matching. They calculate the model curve for trial values of $kh$ , $C$ and $S$ . Then they adjust the values so that the difference between the logarithms of the pressures is as small as possible (advanced section). This is how the match in the lesson's example was found. The principle is the same, but the parameter of the curve changes continuously, and the model can include boundaries, a fracture and heterogeneity of the reservoir.

The manual skill is still useful. It teaches us to see which section of the record sets which parameter, and to notice when a program has “forced” an unsuitable model to fit. Usually we match two curves at once: the pressure and its logarithmic derivative. The lesson Diagnosis with the pressure derivative covers the diagnosis with the derivative.

Terms and symbols

Type curve: A calculated relation between the dimensionless pressure and the dimensionless time for a chosen model. It is usually a family of curves with one parameter, drawn on log–log axes.
Pressure match $(p_D/\Delta p)_M$: The ratio of the dimensionless pressure to the change of pressure at the match point. It gives the flow capacity $kh$ .
Time match $[\Delta t/(t_D/C_D)]_M$: The ratio of the time of the record to the dimensionless time at the match point. It gives the wellbore storage coefficient $C$ .
Match point: Any point whose coordinates we read in both systems after the overlay: on the type-curve plot and on the plot of the record.
Parameter $C_De^{2S}$: The dimensionless wellbore storage coefficient taken at the effective wellbore radius. It sets the shape of the type curve.
Equal scales: The same length of the logarithmic cycle on the plot of the record and on the type-curve plot. Without it, the shape of the curve is distorted.

Common mistakes and misconceptions

Wrong

“The early points coincided, so we have found the right curve.”

In fact

The early points give only $C$ . In the lesson's example, a record 1 h long lay equally well on the curves from $10^5$ to $10^7$ , and the permeability from them differed by a third.

Wrong

“The time match gives the permeability.”

In fact

In the form $p_D$ against $t_D/C_D$ , the time match gives the wellbore storage coefficient $C$ , and the flow capacity $kh$ comes from the pressure match.

Wrong

“We read the skin directly on the curve.”

In fact

On the curve we read the parameter $C_De^{2S}$ . We get the skin as $\tfrac12\ln$ of the ratio of the parameter to $C_D$ . We calculate $C_D$ from the $C$ that we found and from the properties of the reservoir and the well. An error in the wellbore radius or in $\phi c_t$ passes into the skin.

Wrong

“If the record lay well on a type curve, the reservoir model is right.”

In fact

A good fit shows only that the model does not contradict the record. Other models give similar records too. We choose the model by the signs of the whole record and check it against independent data.

Wrong

“We can match a build-up to start-up curves for any producing time before shut-in.”

In fact

Only if the shut-in time is small compared with the producing time. Otherwise the late points of a build-up lie below the start-up curve, and we use the equivalent time (Effect of producing time on build-up analysis).

In depth: advanced detail

How accurate is the single parameter

C_De^{2S}

The curves of the lesson are calculated for a well without skin that has the effective radius $r_we^{-S}$ . The “thin skin” model is an infinitely thin layer with an extra pressure drop at the wall (Skin factor). It is not exactly the same as the effective-radius model. While the disturbance has not moved away from the wellbore, a thin skin and a reduced radius behave differently. A calculation with the same model shows how much this matters. For pairs of $C_D$ and $S$ with the same $C_De^{2S}$ , we compare the $p_D$ of the thin skin with the type curve.

Largest difference between the thin-skin curve and the type curve with the same parameter

$C_De^{2S}$: $10^{2}$
$C_D$: 10
$S$: 1.15
Difference in $p_D$: 1.8%

$C_De^{2S}$: $10^{2}$
$C_D$: 1
$S$: 2.30
Difference in $p_D$: 6.1%

$C_De^{2S}$: $10^{4}$
$C_D$: 100
$S$: 2.30
Difference in $p_D$: 0.17%

$C_De^{2S}$: $10^{4}$
$C_D$: 1
$S$: 4.61
Difference in $p_D$: 2.6%

$C_De^{2S}$: $10^{6}$
$C_D$: 1000
$S$: 3.45
Difference in $p_D$: 0.02%

$C_De^{2S}$: $10^{6}$
$C_D$: 10
$S$: 5.76
Difference in $p_D$: 0.39%

$C_De^{2S}$: $10^{10}$
$C_D$: 100
$S$: 9.21
Difference in $p_D$: 0.04%

$C_De^{2S}$	$C_D$	$S$	Difference in $p_D$
$10^{2}$	10	1.15	1.8%
$10^{2}$	1	2.30	6.1%
$10^{4}$	100	2.30	0.17%
$10^{4}$	1	4.61	2.6%
$10^{6}$	1000	3.45	0.02%
$10^{6}$	10	5.76	0.39%
$10^{10}$	100	9.21	0.04%

Calculation over the interval of $t_D/C_D$ from 0.01 to $10^4$ . The largest difference falls in the transition section, and in radial flow the curves converge.

For $C_D$ from a hundred up, which is almost always the case when a well is shut in at the wellhead, the difference is no more than 0.2%. A record with any pair $C_D$ , $S$ lies on the curve of its parameter. The difference becomes noticeable only for a small $C_D$ , for example when we shut in at the bottom of the well. In such cases the interpretation programs calculate the full model, not one family.

Curves in the coordinates

p_D

against

t_D

The first type curves for a well with wellbore storage and skin were drawn in the coordinates $p_D$ against $t_D$ , with two parameters, $C_D$ and $S$ . On them the early sections are separate unit-slope lines, one for each $C_D$ . The pressure match gives the same flow capacity, and the time match gives the permeability:

Time match on the curves

p_D

against

t_D

k = \phi\mu c_t r_w^2\left(\frac{t_D}{\Delta t}\right)_M

where:

$\left(t_D/\Delta t\right)_M$ dimensionless time of the match point divided by the time of the record that coincided with it, 1/s
$k$ permeability, m²
$\phi,\ \mu,\ c_t,\ r_w$ porosity, viscosity, total compressibility, wellbore radius, —, Pa·s, 1/Pa, m

Units: SI

Holds when

The curves $p_D(t_D)$ are calculated for a known $C_D$ .
The porosity, the compressibility and the wellbore radius are known exactly.

The permeability from the time match and the flow capacity from the pressure match must agree, but comparing them is unreliable. One depends on $\phi c_t$ and on the square of the wellbore radius, and the other depends on the thickness, and these quantities are known inexactly. Besides, the curves for different $C_D$ and $S$ look alike, and it is hard to fit two parameters at once. The form $p_D$ against $t_D/C_D$ brought all the early sections onto one line and left a single parameter, so it is used more often.

In other unit systems (1)

Metric oilfield units

k = \frac{\phi\mu c_t r_w^2}{3.553\cdot10^{-3}}\left(\frac{t_D}{\Delta t}\right)_M

$k$ in mD, $\mu$ in mPa·s, $c_t$ in 1/MPa, $r_w$ in m, $\Delta t$ in h.

$3.553\cdot10^{-3}$ is the factor of dimensionless time in these units.

Matching by fitting: what the program minimises

We can write the overlay by eye as a least-squares problem (find the numbers that make the sum of the squared gaps as small as possible). To match means to find the shifts $a_p = (p_D/\Delta p)_M$ and $a_t = [(t_D/C_D)/\Delta t]_M$ and the parameter $X = C_De^{2S}$ for which the sum of the squares of the differences of the logarithms is smallest:

A measure of the difference between the record and the type curve

F(a_p, a_t, X) = \sum_{i=1}^{n}\left[\ln\left(a_p\,\Delta p_i\right) - \ln p_D\!\left(a_t\,\Delta t_i;\ X\right)\right]^2

where:

$\Delta p_i,\ \Delta t_i$ the measurements: change of pressure and time, MPa, h
$a_p,\ a_t$ shifts along the pressure and time axes: $a_p$ is the pressure match, $a_t$ is the inverse of the time match $[\Delta t/(t_D/C_D)]_M$ , 1/MPa, 1/h
$X$ parameter of the curve $C_De^{2S}$ , —
$p_D(\cdot;\ X)$ the type curve with the parameter $X$ , —
$n$ number of measurements, —

Units: dimensionless

Holds when

All the measurements are equally reliable. When the noise of the gauge is known, we weight the terms.
The model is chosen in advance. The fit does not check that it is right.

Logarithms give the early small and the late large changes of pressure the same weight. This is the same as the overlay on a log–log grid. For given $a_t$ and $X$ , the best pressure shift is found at once: $\ln a_p$ is the mean of the differences $\ln p_D - \ln\Delta p$ . What is left is to fit two numbers. The root of $F/n$ is the root-mean-square relative deviation. In the lesson's example it is about a tenth of a per cent, which is at the level of the rounding of the measurements. A narrow minimum in $X$ means a reliable skin, and a flat minimum means ambiguity, as in the example with the short record. The program will find a minimum in that case too, but the confidence interval of the skin (the range of values that the data allow) will be wide.

Connections to other topics

Wellbore storage — the wellbore storage coefficient, the unit slope and the solution from which the type curves are calculated.
Skin factor — the skin and the effective wellbore radius, on which the parameter $C_De^{2S}$ is built.
Pressure build-up analysis — the Horner method, with which we check the result of matching a build-up.
Pressure drawdown analysis — the semilog straight line for a record after start-up, and its link with the pressure match.
Diagnosis with the pressure derivative — the logarithmic derivative of pressure: diagnosis of the sections of a record, and matching two curves at once.
Nonideal wellbore storage, well clean-up and changing skin — what to do when the wellbore storage coefficient or the skin changes during the record.
Effect of producing time on build-up analysis — how to allow for a short producing time before a build-up.
Dimensionless well solutions: skin, radius of investigation and units — other dimensionless solutions for wells, and how they are written in different units.
Well testing: purposes and types of test — planning the duration of a test so that the record reaches radial flow.

Check yourself

1
What does the pressure match give on the curves $p_D$ against $t_D/C_D$ ?
1. A.The flow capacity $kh$
2. B.The wellbore storage coefficient $C$
3. C.The skin factor
4. D.The porosity of the reservoir
Show the answer
Correct answer: A. The flow capacity $kh$
$p_D = 2\pi kh\,\Delta p/(qB\mu)$ . So the ratio $p_D/\Delta p$ at the match point equals $2\pi kh/(qB\mu)$ , and when the rate and the viscosity are known, we find $kh$ from it. The time match gives $C$ , and the skin comes from the parameter of the curve together with $C_D$ .
2
After matching, the point $p_D = 1$ coincided with $\Delta p$ = 0.40 MPa, and the point $t_D/C_D = 10$ with $\Delta t$ = 0.25 h. The record lay on the curve $C_De^{2S} = 10^4$ . The rate is 50 m³/d, $B$ = 1.1, $\mu$ = 2.0 mPa·s, $h$ = 8 m, $\phi$ = 0.20, $c_t = 1.2\cdot10^{-3}$ MPa⁻¹ and $r_w$ = 0.1 m. Find the permeability, $C$ and the skin.
Show the answer
$qB\mu = 50\cdot1.1\cdot2.0 = 110$ ; $kh = 1.866\cdot110/0.40 \approx 513$ mD·m, $k \approx 64$ mD. $C = 0.02232\cdot(513/2.0)\cdot(0.25/10) \approx 0.143$ m³/MPa. $2\pi\phi c_thr_w^2 = 2\pi\cdot0.20\cdot1.2\cdot10^{-3}\cdot8\cdot0.01 = 1.21\cdot10^{-4}$ m³/MPa, so $C_D \approx 1190$ . $S = \tfrac12\ln(10^4/1190) \approx 1.1$ .
3
A build-up record 2 h long lies entirely on the unit-slope section and the transition section. What can we find reliably from it?
1. A.Only the wellbore storage coefficient $C$
2. B. $C$ and $kh$ , but not the skin
3. C. $kh$ and the skin, but not $C$
4. D.All three parameters
Show the answer
Correct answer: A. Only the wellbore storage coefficient $C$
The unit slope is set only by the wellbore, so $C$ is found with confidence. In the transition section the permeability and the skin compensate each other: neighbouring type curves describe such a record almost equally well, as in the lesson's example with a record 1 h long. For $kh$ and the skin we need radial flow.
4
The parameter of the curve was read with an error of a factor of 2. How much does the calculated skin change?
1. A.By a factor of 2
2. B.It does not change
3. C.By 2
4. D.By 0.35
Show the answer
Correct answer: D. By 0.35
$S = \tfrac12\ln[(C_De^{2S})_M/C_D]$ , so an error of a factor of 2 in the parameter changes the skin by $\tfrac12\ln 2 \approx 0.35$ . The skin is not sensitive to an inexact reading of the parameter, but it is sensitive to $C_D$ , and so to the wellbore radius and to $\phi c_t$ .
5
Why are the porosity, the compressibility of the reservoir and the wellbore radius not in the dimensionless time $t_D/C_D$ , but they are in the skin that we find by matching?
Show the answer
$t_D/C_D = [kt/(\phi\mu c_tr_w^2)]/[C/(2\pi\phi c_thr_w^2)] = 2\pi kh\,t/(\mu C)$ : the $\phi c_tr_w^2$ cancels. So the shifts give $kh$ and $C$ without knowing these quantities. But we get the skin as $\tfrac12\ln[(C_De^{2S})_M/C_D]$ , and $C_D = C/(2\pi\phi c_thr_w^2)$ . Here the porosity, the compressibility, the thickness and the wellbore radius are needed.
6
The pressure at the moment of shut-in is known exactly, but the early points of the record on a log–log plot lie on a line with a slope of about 0.7 instead of 1. What is the most likely cause?
1. A.The skin is too large
2. B.The time count was started later than the actual shut-in of the well
3. C.The time count was started before the shut-in
4. D.The effect of a reservoir boundary
Show the answer
Correct answer: B. The time count was started later than the actual shut-in of the well
On the unit-slope section $\Delta p$ is proportional to the true time. If the clock was started after the shut-in, the recorded times are smaller than the true times by a constant amount. At the start of the record, the ratio of times between the points then looks larger than the ratio of pressures, and the slope is less than 1. A count started before the shut-in gives a slope greater than 1 instead. The skin and the boundaries do not affect this section.

Updated: 5 October 2026